Question

A transition metal $M$ can exist in two oxidation states $+2$ and $+3$. It forms an oxide whose experimental formula is given by $M_{x}O$, where x < 1. Then the ratio of metal ions in $+3$ state to those in $+2$ state in terms of x is given by:

• A. $(1-x)/(1+x)$
• B. 1 + 2x
• C. $1+\frac{x}{2}$
• D. $\frac{2(1-x)}{3x-2}$

Answer 1

The correct option is D $\frac{2(1-x)}{3x-2}$

Given, experimental formula of the metal oxide is $M_{x}O.$
Transition metal $M$ can exist in two oxidation states
$$\begin{array}{ccc}& M^{2+}& M^{3+}\\ \text{Let}& y& (x-y)\end{array}$$
By applying charge balancing.
$2y+3(x-y)-2=0\Rightarrow -y+3x-2=0\Rightarrow y=3x-2$

$\therefore$ Ratio of metal ions,
$\left(\frac{M^{3+}}{M^{2+}}\right)=\frac{x-y}{y}=\frac{x-3x+2}{3x-2}=\frac{2(1-x)}{3x-2}$